*4.11. Water Vapor Diffusivity*

ASTM E96/E 96M-05 method was followed to estimate the Water-Vapor Transmission Rate (WVTR) of ALG/G/xNZ and ALG/G/xTO@NZ films The used apparatus was handmade, and the experiments were carried out at 38 ◦C and 95% RH according to literature [16,23,60,61]. Film samples was of 2.5 cm diameter and 0.09 mm average thickness and placed on the top of an one-open end cylindrical tube made of plexiglass. The cylinder, which contained dried silica gel inside, was sealed by a rubber O-ring. The test tube was placed in a glass desiccator under an environment of 95% relative humidity (RH) at 38 ◦C. Such conditions were obtained placing 200 mL of saturated magnesium nitrate solution in this desiccator. Each tube was weighed periodically for 24 h. The

WVTR (g·cm−<sup>2</sup> ·s −1 ) values was calculated using such weighting measurements and according to Equations (1) and (2):

$$\overline{\left(\frac{\Delta \mathbf{G}}{\Delta \mathbf{t}}\right)} = \frac{1}{\mathbf{n}} \sum\_{i=1}^{n-1} \left(\frac{\mathbf{G}\_{i+1} - \mathbf{G}\_{i}}{\mathbf{t}\_{i+1} - \mathbf{t}\_{i}}\right) \tag{1}$$

$$\text{WVTR} = \overline{\left(\frac{\Delta \mathbf{G}}{\Delta \mathbf{t}}\right)} \cdot \frac{1}{\mathbf{A}} \tag{2}$$

where: ∆G ∆t (g·s −1 ) is the mean water vapor mass diffusion rate for the overall experiment duration, G<sup>i</sup> (g) is the device weight which balanced at the elapsed time t<sup>i</sup> (s) from the experiment start, <sup>G</sup>i+1−G<sup>i</sup> ti+1−t<sup>i</sup> (g·s −1 ) is the water vapour mass diffusion rate for each time interval, (n) is the number of weight measurements, and A (cm<sup>2</sup> ) is the permeation area of the film. Additionally, the tested films were weighed before and after the WVTR test to exclude any absorption phenomena of humidity in the film. ∆G/∆t (g/s) is the water transmission rate through the film which is calculated by the experimental points (G<sup>i</sup> , ti ).

WVTR is equal to the specific mass flow rate for the diffusion process through a membrane which could be calculated using Fick's law [62]:

$$\frac{\text{J}}{\text{A}} = \text{D} \cdot \frac{\Delta \text{C}}{\Delta \text{x}} \tag{3}$$

where J (g/s) is the mass flow rate of a component through the membrane, A (cm<sup>2</sup> ) is the membrane cross-sectional area permeated by this component, <sup>∆</sup>C (g·cm−<sup>3</sup> ) is the concentration gradient of this component on the two sides of the membrane, and ∆x (cm) is the membrane thickness. the humidity concentration in the outer side of the cylinder is 4.36509 <sup>×</sup> <sup>10</sup>−<sup>5</sup> <sup>g</sup>·cm−<sup>3</sup> (95% RH at 38 ◦C) and in the opposite side of the film the silica gel absorbs the 100% of the permeated water vapor, which agrees with the ASTM E96/ E 96M-05 method, then <sup>∆</sup>C = 4.36509 <sup>×</sup> <sup>10</sup>−<sup>5</sup> <sup>g</sup>·cm−<sup>3</sup> . Considering that WVTR = J/A we can calculate the diffusion coefficient D (cm<sup>2</sup> ·s −1 ) for every film via the combination of Equations (2) and (3) which leads to Equation (4):

$$\mathbf{D}\_{\rm WV} = \mathbf{W} \mathbf{V} \mathbf{T} \mathbf{R} \cdot \frac{\Delta \mathbf{x}}{\Delta \mathbf{C}} \tag{4}$$

where WVTR [g·cm−<sup>2</sup> ·s −1 )] is the water-vapour transmission rate, ∆x (cm) is the film thickness, and <sup>∆</sup>C (g·cm−<sup>3</sup> ) is the humidity concentration gradient on the two opposite sides of the film.
